A gambler bets $1 on a fair coin, doubles the stake after each loss and stops at the first win. His profit is $1 with probability 1, yet his wealth is a martingale starting at 0. Which condition of the optional stopping theorem fails, and what does the strategy cost in practice?

A gambler bets $1 on a fair coin, doubles the stake after each loss and stops at the first win. His profit is $1 with probability 1, yet his wealth is a martingale starting at 0. Which condition of the optional stopping theorem fails, and what does the strategy cost in practice?

Approach: Check each sufficient condition of the theorem in turn against this stopping time and this process, then compute the distribution of the loss carried just before the winning bet.

Uniform integrability fails, and with it the conclusion of the optional stopping theorem. The theorem holds when the stopping time is bounded, when it is finite with bounded increments, or when the stopped martingale is uniformly integrable. Here T is geometric with E[T] = 2, so it is finite, but the k-th stake is 2^{k-1} and the increments are unbounded, so the first two conditions fail. The drawdown carried into the winning bet on trial k is 2^{k-1} - 1, and E[drawdown] = sum_k 2^{-k}(2^{k-1} - 1) diverges, so no finite bankroll supports the strategy. Truncating at a bankroll B after log_2(B) losses restores the theorem exactly: the truncated version has expected profit 0, with many $1 wins offset by a rare loss of about B.

Follow-up: What is the probability of being down at least $1000 at some point before the first win?

Key concepts: optional stopping theorem, uniform integrability, martingale, drawdown.